Part A The lines show the equipotential contours in the plane of three point cha
ID: 1575290 • Letter: P
Question
Part A The lines show the equipotential contours in the plane of three point charges, Q. Q2, and Q. The values of the potentials are in kV as indicated for the +5, 0, and -5 kV contours. The positions of the charges are indicated by the dots. Cm The letters are on the equipotential contours. False: The force on a proton ato points to the bottom of the page. True: The electric field ati is stronger than atj. False: The electric field at k is zero. False: Charge Q2 is the largest negative charge. True: Q is a negative charge. True: Charge Q; has the largest magnitude of all. You are correct. Your receipt no. is 162-340 o Previous Tries Part B Calculate the work required to move a charge of -0.47x10-12C from i to b. 2.82x10-9 ) You are correct. Your receipt no. is 162-1815 Previous Tries Part C Calculate the size of the electric field at k. 1.333x10A.O NUC The voltage decrease divided by the distance along which the voltage decreases the most. | Submit Answer Incorrect. Tries 3/15 Previous Tries Part D Calculate the size of the force on a charge of 8.00x10-19C at g. Submit Answer Tries 0/15 Part E Calculate the size of Q. The magnitudes of the three charges are in the exact ratios of 1 to 2 to 3.Explanation / Answer
Ans:-
Part c :
The perpendicular distance between the equipotentials either side of g (-1kV and +1kV) is 12.5mm.
(to measure distance XY, place the straight edge of a sheet of paper against X and Y, mark the positions of X and Y on the edge of the paper, and measure this distance using the scale given in the figure.)
electric field strength at g = PD/distance = 2x10^3V/12.5x10^-3mm = 0.16x10^6 V/m.
Part d
the perpendicular distance between the equipotentials either side of k (-1kV and +1kV) is 9mm.
electric field strength at k = PD/distance = 2x10^3V/9x10^-3mm = 0.22x10^6 V/m.
force on charge at k = charge x electric field strength = 8×10^-19 C x 0.22x10^6 V/m = 1.76x10^-13 N.
Part e
(THE DIFFICULT PART.) to find the charge Q3 we can use the SUPERPOSITION PRINCIPLE : the electrostatic potential at any point in the diagram is the sum of the potentials from all 3 charges. the potential at distance r from a charge Q is Q/4r, where is the permittivity of free space. we pick any convenient point X on an equipotential and measure the distances r1, r2,r3 of Q1, Q2, Q3 from X. then
(1/ 4)(Q1/r1 + Q2/r2 + Q3/r3) = P
where P = potential at X.
(comment: if required, we could use this equation to find all 3 charges Q1, Q2, Q3 by selecting 3 different points X, Y, Z then solving 3 simultaneous equations. alternatively, we could measure distances from several points and average the resulting values of Q3 to improve accuracy.)
judging by the distance from each charge to the nearest equipotential (+5 or -5 kV), Q3 has the largest magnitude (because Q/r is the same for each charge, so large distance r to the equipotential means Q must also be large) and is +ve. Q1 is comes in between and is -ve, and Q2 is smallest, also -ve.
so Q3 : Q1 :Q2 = +3 : -2 : -1. if Q2 = -q then Q1 = -2q and Q3 = +3q. substituting above,
q(-2/r1 -1/r2 + 3/r3) = 4P.
for accuracy, choose X as far from each charge as possible. X cannot be on the P= 0 V equipotential, because the equation would then give us q=0 for all values of r. i choose X as the point where the +1kV equipotential meets the left lower edge.
distances measured to X from Q1, Q2, Q3 are 62, 70, 35.5 mm respectively.
P= +1kV, = 8.85x10^-12 F/m, 4P = 1.1x10^-7 C/m.
(-2/r1 -1/r2 + 3/r3) = (-2/0.062 -1/0.070 + 3/0.0355) = 37.96 m^-1.
q = 1.1x10^-7 / 37.96 = 2.92 x 10^-9 C.
ANSWER: Q3 = 3x (+2.92nC) = +8.8 nC.